The centre and radius of the circle (a + 1)x^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the equation to represent a circle, the coefficients of $x^2$ and $y^2$ must be equal. Thus, $a + 1 = 3$, which gives $a = 2$.Substituting $a

Vedclass · Accepted Answer

$(1, -3/2), \sqrt{5}/2$

Vedclass · Answer

(C) For the equation to represent a circle, the coefficients of $x^2$ and $y^2$ must be equal. Thus, $a + 1 = 3$, which gives $a = 2$.Substituting $a = 2$ into the equation: $3x^2 + 3y^2 - 6x + 9y + 6 = 0$.Dividing by $3$: $x^2 + y^2 - 2x + 3y + 2 = 0$.Comparing with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, we get $2g = -2 \implies g = -1$ and $2f = 3 \implies f = 3/2$.The centre is $(-g, -f) = (1, -3/2)$.The radius is $\sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (3/2)^2 - 2} = \sqrt{1 + 9/4 - 2} = \sqrt{9/4 - 1} = \sqrt{5/4} = \sqrt{5}/2$.

The centre and radius of the circle $(a + 1)x^2 + 3y^2 - 6x + 9y + a + 4 = 0$ are respectively ...

Similar Questions

The radius of the circle $r=12 \cos \theta+5 \sin \theta$ is

The abscissae of two points $A$ and $B$ are the roots of the equation $x^2+2ax-b^2=0$ and their ordinates are roots of the equation $y^2+2py-q^2=0$. Then the equation of the circle with $AB$ as diameter is given by

The number of integral values of $K$ for which $x^2+y^2+kx+(1-k)y+5=0$ represents a circle whose radius cannot exceed $5$ is:

If $P_1, P_2, P_3$ are the perimeters of the three circles $x^2+y^2+8x-6y=0$,$4x^2+4y^2-4x-12y-186=0$,and $x^2+y^2-6x+6y-9=0$ respectively,then

The equation of the circle with centre $(2,3)$ and touching the line $3x-4y+1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module